Ex 5.1, 11 - Find all points of discontinuity f(x) = { x3 - 3 - Ex 5.1

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Ex 5.1, 11 Find all points of discontinuity of f, where f is defined by ๐‘“ ๐‘ฅ๏ทฏ= ๐‘ฅ3โˆ’3, ๐‘–๐‘“ ๐‘ฅโ‰ค2๏ทฎ&๐‘ฅ2+1 , ๐‘–๐‘“ ๐‘ฅ>2๏ทฏ๏ทฏ We have, ๐‘“ ๐‘ฅ๏ทฏ= ๐‘ฅ3โˆ’3, ๐‘–๐‘“ ๐‘ฅโ‰ค2๏ทฎ&๐‘ฅ2+1 , ๐‘–๐‘“ ๐‘ฅ>2๏ทฏ๏ทฏ Case 1: At x = 2 f is continuous at x = 2 if L.H.L = R.H.L = ๐‘“ 2๏ทฏ i.e. lim๏ทฎxโ†’ 2๏ทฎโˆ’๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = lim๏ทฎxโ†’ 2๏ทฎ+๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = ๐‘“ 2๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = ๐‘ฅ3โˆ’3 ๐‘“ 2๏ทฏ = 2๏ทฏ3โˆ’3 = 8 โˆ’ 3 = 5 Thus LHL = RHL = f (2) โ‡’ f is continuous at ๐’™=๐Ÿ Case 2 Let x = c , where c < 2 ๐‘“ ๐‘ฅ๏ทฏ=๐‘ฅ3โˆ’3 f is continuous at x = c if lim๏ทฎxโ†’๐‘๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ=๐‘“(๐‘) Thus, lim๏ทฎxโ†’๐‘๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ=๐‘“(๐‘) โ‡’ f is continuous for ๐‘ฅ =๐‘ less than 2. โ‡’ f is at continuous for all real numbers less than 2. Case 3 Let x = c (where c > 2) ๐‘“ ๐‘ฅ๏ทฏ=๐‘ฅ2+1 f is continuous at x = c if lim๏ทฎxโ†’๐‘๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ=๐‘“(๐‘) Thus lim๏ทฎxโ†’๐‘๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ=๐‘“(๐‘) โ‡’ f is continuous at ๐‘ฅ =๐‘ ( c is greater than 2) โ‡’ f is continuous at all real numbers greater than 2. Hence, there is no point of discontinuity โ‡’ f is continuous at all real point. Thus, f is continuous for all ๐’™โˆˆ๐‘.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.